1.

E^(@)(Cu^(2+)//Cu) and E^(@)(Ag^(+)//AG)" is "0.337V and +0.799 V respectively. Make a cell whose EMF is +ve. If the concentration of Cu^(2+)" is 0.01M and "E_("cell")" at "25^(@)C is zero, calculate the concentration of Ag^(+).

Answer»

Solution :Cu is more reactive than silver, so that the cell is as `Cu//Cu_(2)^(+) (0.01M) || Ag^(+) (C )//Ag` or cell reaction
`Cu+2Ag^(+)rarrCu^(2+)+2Ag`
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)log.([Cu^(2+)][Ag]^(2))/([Cu][Ag^(+)]^(2))`
`=E_("cell")^(@)-(0.0591)/(n)log.((0.01)xx1^(2))/(1xx[Ag^(+)]^(2))`
`"Or"[Ag^(+)]=1.47xx10^(-9)M`


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